Question:
The set
\[
\left\{ \frac{1}{m} + \frac{1}{n} : m, n \in \mathbb{N} \right\} \cup \{0\}, \text{ as a subset of } \mathbb{R}, \text{ is}
\]
The set
\[
\left\{ \frac{1}{m} + \frac{1}{n} : m, n \in \mathbb{N} \right\} \cup \{0\}, \text{ as a subset of } \mathbb{R}, \text{ is}
\]
Show Hint
A set is compact if it is closed and bounded, and it is open if it contains none of its boundary points.
Updated On: Dec 11, 2025
- compact and open
- compact but not open
- not compact but open
- neither compact nor open
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the set.
The set \( \left\{ \frac{1}{m} + \frac{1}{n} : m, n \in \mathbb{N} \right\} \cup \{0\} \) consists of sums of reciprocals of natural numbers, along with \( 0 \) as a limit point. As \( m \) and \( n \) become large, the terms approach 0, but the set does not include any points greater than 0.
Step 2: Analyzing compactness and openness.
The set is compact because it is closed and bounded. The set includes its limit point 0, and all points are contained within a bounded region of the real line. However, the set is not open, as there are no intervals of real numbers fully contained in the set.
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(B)} \).
The set \( \left\{ \frac{1}{m} + \frac{1}{n} : m, n \in \mathbb{N} \right\} \cup \{0\} \) consists of sums of reciprocals of natural numbers, along with \( 0 \) as a limit point. As \( m \) and \( n \) become large, the terms approach 0, but the set does not include any points greater than 0.
Step 2: Analyzing compactness and openness.
The set is compact because it is closed and bounded. The set includes its limit point 0, and all points are contained within a bounded region of the real line. However, the set is not open, as there are no intervals of real numbers fully contained in the set.
Step 3: Conclusion.
Thus, the correct answer is \( \boxed{(B)} \).
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